Problem: The sides of rectangle $ABCD$ have lengths $10$ and $11$. An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$.  Find the maximum possible area of such a triangle.
Solution: Place the rectangle in the complex plane so that one corner is at the origin, and the sides align with the real and imaginary axis.  To maximize the area of the triangle, we let one vertex of the triangle be at the origin, and we let the other two vertices ($p$ and $q$) lie on the sides of the rectangle, as shown.

[asy]
unitsize(0.4 cm);

pair A, B, C, D, P, Q;

A = (0,0);
B = (11,0);
C = (11,10);
D = (0,10);
Q = extension(C, D, rotate(60)*(B), rotate(60)*(C));
P = rotate(-60)*(Q);

draw(A--B--C--D--cycle);
draw(A--P--Q--cycle);

label("$0$", A, SW);
label("$p$", P, E);
label("$q$", Q, N);
label("$11$", B, SE);
label("$10i$", D, NW);
[/asy]

Then $p = 11 + yi$ for some real number $y.$  Also,
\begin{align*}
q &= e^{\pi i/3} p \\
&= \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) (11 + yi) \\
&= \left( \frac{11}{2} - \frac{\sqrt{3}}{2} y \right) + i \left( \frac{y}{2} + \frac{11 \sqrt{3}}{2} \right).
\end{align*}Since the imaginary part of $q$ is 10,
\[\frac{y}{2} + \frac{11 \sqrt{3}}{2} = 10,\]so $y = 20 - 11 \sqrt{3}.$

Then the area of the triangle is
\begin{align*}
\frac{\sqrt{3}}{4} \left|11 + (20 - 11 \sqrt{3}) i\right|^2 &= \frac{\sqrt{3}}{4} \left(11^2 + (20 - 11 \sqrt{3})^2\right) \\
&= \frac{\sqrt{3}}{4} (884 - 440 \sqrt{3}) \\
&= \boxed{221 \sqrt{3} - 330}.
\end{align*}